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Operation research bba15

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Operation research bba15

  1. 1. Presented by :- Sonal chauhan TOPIC OF PRESENTATION PRESENTED TO:- Bba 2ND YR
  2. 2. INTRODUCTION Before understanding the meaning of linear programming, its beneficial to understand the words linear and programming individually. So, LINEAR means that the relationships are represented by straight lines i.e., the relationships are in the form of y =ax+b. PROGRAMMING means the optimal allocation of resources. Hence, linear programming can be defined as mathematical technique for determining the optimal allocation of resources and obtaining a particular objective by making choices from the available sources .The objective may be cost minimization or inversely
  3. 3. APPLICATION KINDS OF L.P.P. There are some important L.P problem discussed below:-  Optimal product line problems : Production and sale of different products by the manufacturer to make maximum profit.  Product mix problems: Keeping in mind the different conditions we try to determine such a product mix which maximizes the profit and minimizes the cost.  Diet planning problems: We have to determine the amount of different kinds of nutrients which should be included in a diet so as to minimize the cost .  Transportation problems: Taking an example ,the finished goods from the plant are also to be transported to warehouse in such a way that the total
  4. 4.  A set of values of variables satisfying the constraints of a L.P.P is called a solution of a L.P.P. Say for example:- lets take an equation (2x + 3y = 10) here CONSTRAINT VARIABLE Similarly 3 is constraint and Y is variable CONSTRAINTS can be changed but VARIABLES remains constant.  The variable used in linear programming problem to convert an inequality in to an equation is known as slack variable. FOR EXAMPLE- 2x + 3y ≤ 100 , 3x +4y ≤ 200 These are known as inequality and for converting and for solving them we assume some slack variables which are always equal to 0 and which does not lays any effect on the given inequality and converts them in a equation. As we consider some slack variables as S1, S2, =0 .
  5. 5. 2x + 3y + S1 = 100 3x + 4y + S2 = 200 Next we will put the value of S1 and S2 i.e., 0 and finally we will get the equations as-: 2X +3y = 100 3x + 4y = 200 which can be solved further. • General form of L.P.P is Maximization or minimization Z= a₁x₁ + a₂x₂…..anxn subject to :- a₁₁x₁ + a₁₂x₂+ a₁₃x₃……..≤b₁ , a₂₁x₁ + a₂₂x₂ + ……≤b₂…. For Example MAXIMIZE Z=25x₁ + 40x₂ MINIMIZE Z=2x₁ + 3x₂ Subject to : x₁ + 2x₂ ≤ 5 Subject to : 3x₁ + 2x₂ ≥ 5 2x₁ + 5x₂ ≤ 6 2x₁ + 5x₂ ≥ 7
  6. 6. METHODS OF SOLVING LINEAR PROGRAMMING PROBLEMS GRAPHICAL METHOD SIMPLEX METHOD (Here we will discuss only graphical method in detail)
  7. 7. GRAPHICAL METHOD For a L.P.P, that have only two variables ,it is possible that entire solution can be displayed graphically by plotting linear constraints on a graph paper to locate best solution. This technique is called GRAPHICAL METHOD.  It is applicable when only two variables are involved.  There are two types of graphs i.e., maximization in case of profit and minimization in case of cost. Let’s take an example to understand the graphical method practically and more easily.Maximize Z = f(x,y) = 3x + 2y subject to: 2x + y ≤ 18 2x + 3y ≤ 42 3x + y ≤ 24 x ≥ 0 , y ≥ 0
  8. 8. 1. Initially ,the coordinate system is drawn and each variable is associated to an axis. 2. A numerical scale is marked in axis. 3. In the beginning ,for changing the given inequalities into an equation we will consider some slack variables ,say( S1,S2,S3=0) and add one slack variable in each inequality respectively. 2x + y + S1 = 18 , 2x + 3y +S2 = 42, 3x + y +S3 = 24 Now put the value of S1, S2, S3 =0 4. After putting the value of S1, S2, S3 in the above given equations we will get the following results :- 2x + y + S1 = 18 2x + y + 0 = 18 (a) 2x + 3y + S2 = 42 2x + 3y + 0 = 42 (b) 3x + y + S3 = 24 3x + y + 0 = 24 (c) 5. Next we will find out the coordinates from the given above equation (a) , (b) , (c) by taking each variable x and y one by one as 0.
  9. 9. Taking equation (a) for finding out the coordinates . If x=0, then 2x + y = 18 will be 2 X 0 + y = 18 0 + y = 18 y = 18 If y =0 , then 2x + y = 18 will be 2x+ 0 = 18 2x = 18 x = 9, therefore the coordinates we get will be (9,18) Taking equation (b) for finding out the coordinates. If x = 0, then 2x + 3y = 42 will be 2 X 0 + 3y = 42 0 + 3y = 42 y = 14 If y = 0 , then 2x + 3y = 42 will be 2x + 3 X 0 = 42 2x + 0 = 42 x = 21, therefore the coordinates we get will be ( 21 , 14 )
  10. 10. Taking equation (c) for finding out the coordinates . If x=0, then 3x + y = 24 will be 3 X 0 + y = 24 0 + y = 24 y = 24 If y =0 , then 3x + y = 24 will be 3x+ 0 = 24 3x = 24 x = 8, therefore the coordinates we get will be (8,24) 6. As we have got the coordinates (9,18) , (21,14) , (8,24) from the equation (a) , (b) , (c) respectively . We will plot these coordinates on the graph . 7. We will plot the coordinates (9,18) first as shown in the figure by drawing straight line joining two points 9 on the x axis and 18 on y axis.
  11. 11. 8. Next we will plot the coordinates (21,14) On the graph shown beside. 9. Further we will plot the coordinates (8,24) Shown here with the red shaded region.
  12. 12. Since we have taken the example of maximization therefore the region towards the origin will be covered.
  13. 13. 10. The required region is the intersection of the regions defined by the set of constraints and the coordinate axis . This required region is represented by the O-F-H-G-C polygon in PURPLE color. 11. Next, the extreme values are calculated. These vertices are the optimal solutions. 12. For point O , C and F the coordinates will be (0,0) , (0,14) , (8,0)
  14. 14. 13. For point G, we will solve the equations (a) and (b) , and after solving these equations we will get the coordinates (3,12). Similarly for point H, we will solve the equations (a) and (c) and we will get the coordinates as (6,6). 14. 15. Finally, the objective function (3x + 2y) is evaluated in each of these points (results are shown in the table above). Since G-point provides the greatest value to the Z- function and the objective is to maximize, this point is the optimal solution: Z = 33 with x = 3 and y= 12. Extreme point Z = 3x + 2y Objective value (Z) O(0,0) Z = 0 + 0 0 C(0,14) Z = 0 + 28 28 G(3,12) Z = 9 + 24 33 H(6,6) Z = 18 + 12 30 F(8,0) Z = 24 + 0 24
  15. 15. ANY QUESTION ?

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